The eigenvectors of a matrix are those special vectors
for which
, where is an associated
constant (possibly complex) called the eigenvalue. Let us
rearrange the eigenvalue equation
to the form
, where
represents a vector of all zeroes (the zero vector). We may rewrite this
expression using the identity matrix to yield
, which will be more convienient for the next
step. Now to solve for , multiply the left and right sides
of the equation by
, if it exists. This
yields
.

This last equation presents a challenge. Anything multiplied by the zero
vector yields the zero vector, but clearly that is a trivial solution
for that we aren't interested in. Thus, our assumption that
exists must be wrong. The eigenvalue
equation
therefore has non-trivial solutions
only when
does not exist. In
our previous discussion of determinants, we noted that a matrix
does not have an inverse if its determinant is zero. Thus, we
can satisfy the eigenvalue equation for those special values of
such that
. This is called the
secular determinant, and expanding the determinant gives
an -th degree polynomial in called the secular equation
or the characteristic equation. Once the roots of this equation are
determined to give eigenvalues , these eigenvalues may be
inserted into the eigenvalue equation, one at a time, to yield
eigenvectors.

As an example, let us find the eigenvalues and eigenvectors for the matrix

(25)

We begin with the secular determinant
,
which in this case becomes

(26)

Expanding out this determinant using the rules given above for the
determinants of matrices, we obtain the following
characteristic equation:

(27)

which has solutions
. These are the three eigenvalues
of . What are the corresponsing eigenvectors? We substitute each
of these eigenvalues, one at a time, into the eigenvalue equation, and
solve for the system of equations that result.

Let us begin with the eigenvalue . Substituting this into
, we obtain

(28)

This is three equations in three unknowns, which we may rewrite as

(29)

(30)

(31)

The middle equation is, of course, not particularly useful. However,
we know that the components and of the eigenvector
corresponding to are both zero, and there is no equation
governing the choice of . We are therefore free to chose any
value for , and a valid eigenvector will result. Note that any
eigenvector times a constant will yield another valid eigenvector.
Most frequently, we chose normalized eigenvectors by convention
(such that
), so in this case we will choose
. This gives the final eigenvector

(32)

We can verify that this is indeed an eigenvector corresponding to the
eigenvalue by multiplying this eigenvector by the original
matrix :

(33)

that is, multiplication of times the eigenvector yields the
eigenvector again times a constant (the eigenvalue, ).

By a similar procedure, one can obtain the other two eigenvectors, which,
when normalized, are

(34)

Although an matrix has eigenvalues, they are
not necessarily distinct. That is, one or more of the roots of
the characteristic equation may be identical. In this case, we say that
those eigenvalues are degenerate. Determination of eigenvectors is
somewhat more complicated in such a case, because there will be additional
flexibility in selecting them. Consider the matrix

(35)

which has the characteristic equation

(36)

with solutions
. Although it is no difficulty to find
the eigenvector corresponding to , the doubly-degenerate
eigenvalue presents an additional complication. Upon
substituting this value into the eigenvalue equation, we obtain

(37)

or

(38)

In this case, the first and third equations are equivalent (one equation is
just the minus of the other), and so they are not independent.
Additionally, the second equation gives no information. We therefore
have only one equation to determine the three coefficients of the
eigenvector. Recall that only two valid equations resulted above for our
previous non-degenerate case, because any multiple of an eigenvector still
yields a valid eigenvector. Here, a double degeneracy has lost us one of
our equations. Hence, we only know that and is
arbitrary. Any eigenvector satisfying these rules will be satisfactory,
and clearly there are an infinite number of ways to chose them, even if we
require them to be normalized. Let us, somewhat arbitrarily, pick the
normalized vector

(39)

Then, it is traditional to try to pick the second eigenvector for
as orthogonal to the first (there are reasons for doing this,
most commonly because we might wish to use these vectors as a new
orthonormal basis). In that case, the following (normalized) vector will
be suitable:

(40)

You can verify that both of these vectors are (a) orthonormal, and (b)
satisfy the eigenvalue equation for .